Resultant Velocities / Elastic Collision / 1-D?

[julianwitkowski]

[Moderator's note: No template because the thread had attracted a fair number of responses before being moved out of Classical]

Hey it's my first post, so I apologize in advance if this is under the wrong forum category... I am just wondering if I did I do this right? If not can you please show me what I've done wrong and show your corrected math.

Thank you!

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E].

What is the final velocity of each cart after the collision?

First I changed the frame of reference...

FROM [0.8 kg @ 2 m/s]----> <----[0.6 kg @ -5 m/s]

TO [0.8 kg @ 7 m/s]----> <----[0.6 kg @ 0 m/s] (are these equations right?)


My work...

v₁'=[(m₁-m₂)/(m₁+m₂)]*v₁

v₁'=[(0.8-0.6)/(0.8+0.6)]*7

v₁'=1 v₂'=[(2* m₁)/(m₁+m₂)]*v₁

v₂'=[(2* 0.8)/(0.8+0.6)]*7

v₂'=8 Final Velocities

vₓ₁= v₁' + -5

vₓ₁= 1 + -5

vₓ₁= -4 m/s vₓ₂= v₂' + -5

vₓ₂= 8 + -5

vₓ₂= 3 m/s The final velocity for the 0.8 kg is 4 m/s [W]

The final velocity for the 0.6 kg is 3 m/s [E]

 

[Bystander]

Elastic collision: what all is conserved?

 

[julianwitkowski]

Elastic collision: what all is conserved?

The collision is cushioned by a spring (k = 1200 N/m), but that's only necessary for the second question which is "What is the max compression of the spring?"

 

[Bystander]

What physical quantities are conserved in an elastic collision?

 

[julianwitkowski]

What physical quantities are conserved in an elastic collision?

Can you please tell me? I don't know...

 

[Bystander]

You jest. Linear momentum is conserved. Kinetic energy is conserved. You have covered those two concepts, have you not?

 

[julianwitkowski]

You jest. Linear momentum is conserved. Kinetic energy is conserved. You have covered those two concepts, have you not?

I don't know, I'm not really good at physics, but that's what the textbook says to do; however, someone asked about this problem on Yahoo answers, and my answer is different from the person who answered it (who is highly ranked in the physics answers section)... I think if I got anything wrong its going back to the ground observer's frame of reference.

 

[Bystander]

One step at a time. What is the linear momentum of the westbound cart? What is the linear momentum of the eastbound cart? Calculate those two values for me, and do NOT shift reference frames, it ain't necessary.

 

[julianwitkowski]

One step at a time. What is the linear momentum of the westbound cart? What is the linear momentum of the eastbound cart? Calculate those two values for me, and do NOT shift reference frames, it ain't necessary.

Do you mean like this?

east
p = m*v = 0.8 kg * 2 m/s = 1.6 N/s

west
p = m*v = 0.6 kg * -5 m/s = -3 N/s

 

[Bystander]

Close. Instead of N/s you'll get units of linear momentum, mass x velocity, or kg⋅m/s. Numbers are good. Now, what is the total linear momentum of the system (that's both carts)? And, we can start on step 2: what is the kinetic energy of westbound? What is kinetic energy of eastbound?

 

[julianwitkowski]

Close. Instead of N/s you'll get units of linear momentum, mass x velocity, or kg⋅m/s. Numbers are good. Now, what is the total linear momentum of the system (that's both carts)? And, we can start on step 2: what is the kinetic energy of westbound? What is kinetic energy of eastbound?

p = (m ⋅v) + (m ⋅v) = -1.4 kg ⋅m/s

west ke = ½ m ⋅v² = ½ 0.8 ⋅2² = 1.6 J
east ke = ½ m ⋅v² = ½ 0.6 ⋅-5² = 7.5 J

total ke = ( ½ m ⋅v²)+( ½ m ⋅v²) = 9.1 J

 

[Bystander]

Good. You now have two equations in two unknowns, v_E and v_W. Solve the total momentum for either in terms of the other, substitute that into the total kinetic energy, and you'll get a quadratic which you can solve. You'll get two roots, and one should be obviously useless, take the other back to total momentum and solve for both velocities.

 

[julianwitkowski]

Good. You now have two equations in two unknowns, v_E and v_W. Solve the total momentum for either in terms of the other, substitute that into the total kinetic energy, and you'll get a quadratic which you can solve. You'll get two roots, and one should be obviously useless, take the other back to total momentum and solve for both velocities.

I read my textbook more, and it said, the whole reason you change the frame of reference is to avoid doing all this work, because when you simplify the algebra it comes out to the equations I showed in my question... So really, all I wanted was to see if you got the same answer yourself doing it that I did... The Ke and momentum are of no consequence because they will divide out.